ZEROES OF INTEGER LINEAR RECURRENCES. 1. Introduction. 4 ( )( 2 1) n

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1 ZEROES OF INTEGER LINEAR RECURRENCES DANIEL LITT Consider the integer linear recurrence 1. Introduction x n = x n 1 + 2x n 2 + 3x n 3 with x 0 = x 1 = x 2 = 1. For which n is x n = 0? Answer: x n is never zero, because it s always positive. What about x n = 2x n 1 + x n 2 with x 0 = 2, x 1 = 1? Answer: Only n = 3, because ( x n = 1 3 ) 2 ( 1 2) n ( )( 2 1) n and the left term dominates for large n. (Discuss when this will wor, drawing roots of polynomial.) What about x n = 2x n 1 3x n 2 with x 0 = 0, x 1 = 1? Answer: Only x 0 ; loo mod 3. (Absolute value doesn t wor, because roots are 1 ± i 2 which have the same absolute value.) Mod 3 this is ( 1) n+1. Finally, how about x n = x n 2, with x 0 = 0, x 1 = 1? Answer: Even n, obviously. The goal of this tal is to show that these sorts of behaviors are the only thing that can happen really; we ll prove a beautiful theorem of Solem-Mahler-Lech to that effect. Theorem 1 (Solem-Mahler-Lech). Let a n be a sequence defined by an integer linear recurrence. Then the set A := {n a n = 0} is the union of a finite set and a finite collection of (right)-infinite arithemtic progressions, i.e. sets of the form {qn + r n N}. 1

2 2. The p-adics Let s loo at the example more closely. We have x n = 2x n 1 3x n 2 x n = (1 2) n (1 + 2) n. I ve purposefully written this so it maes sense mod 3. Now 2 = ±1 mod 3, giving the mod 3 identity I claimed above. This is rigorous because 2 has a square root mod 3; of course looing mod 3 might not have wored; we might have had to loo mod a higher power; for example, if our initial conditions were x 0 = 0, x 1 = 3. But that s o we can lift our roots mod 9. Namely, pic an arbitrary lift of one of our roots, say 1; now we want to write (3a + 1) 2 = 2 mod 9, or 6a + 1 = 2 mod 9, which has a solution (namely a = 1). Why does this wor? Well in general, lets say we have a root x of some polynomial p mod 3. We can pic some lift x mod 9 and try to fix it up, e.g. we want p(3a + x) = 0 mod 9 for a {0, 1, 2}. In other words 3p ( x)a + p( x) = 0 mod 9 But p( x) = 0 mod 3 by our choice of x, so a solution exists for a if p ( x) is invertible; that is, if p (x) 0. By the way, this is just Newton s method! Of course, nothing is special about 3 and 9, so we ve really just proved Lemma 1 (Hensel s Lemma). Let f be a polynomial with integer coefficients and p a prime. Suppose x is a root of f mod p, and f (x) 0. Then the exists x Z/p Z so that f(x) = 0 and x = x mod p. Proof. Induction on. I claim this method of analyzing linear recurrences looing mod large powers of primes is secretly the same as the previous method using absolute values. Namely, let s consider this cleaner statement of Hensel s lemma. Define the p-adic integers Z p := lim Z/p Z. An element of Z p is by definition a sequence of elements a Z/p Z so that a = a 1 mod p 1. Then Hensel s Lemma says that if a polynomial f with integer coefficients (or really with coefficients in Z p!) has a x F p with f (x) 0, then f has a solution x in Z p so that x = x mod p. Z p is a topological ring, with multiplication inherited from the finite levels and a topology induced by giving each Z/p Z the discrete topology. (So open sets are generated by preimages of subsets of Z/p Z for some.) [Draw a picture of Z 3 ; note that any point in a ball is its center] Now each p-adic number has a decimal expansion z = 0 a p where p {0, 1, 2,, p 1} by the explicit description of the inverse limit; note that those numbers with finite decimal expansions are just the usual integers. Just lie the usual decimal expansion of a number, this should suggest another description of the p-adics. Namely, if z is a non-zero integer, define Let v p (z) = max{ N p divides z}. z p = p vp(z) for z nonzero, and let 0 p = 0. So for example 24 2 = 2 3 = 1/8. This is an absolute value in the sense that it satisfies: 0 p = 0; x p y p = xy p ; x p + y p x + y p ; indeed max( x p, y p ) x + y p. (This is called the ultrametric inequality). 2

3 This absolute value induces a metric on Z; its completion with respect to the induced topology is Z p! The extension of v p to Z p is that v p ( a p ) = min( a 0). Alright, so let s go bac to our example x n = 2x n 1 3x n 2. The formula x n = (1 2) n (1 + 2) n. maes sense in Z 3 by Hensel s lemma, where we tae 2 = 1 mod 3. But now so we may apply our old absolute value argument! = 1, and = 1/3 3. Topology and Analysis in the p-adics Let s establish some properties of the p-adics. I ve already asserted their completeness; one way to see this is that if b is a Cauchy sequence, the i-th digit of b stabilizes (why?), giving the i-th digit of the limit. Having convergence of sequences, let us move on to convergent series. Lemma 2. Let b be a sequence of p-adic integers. Then b converges iff b p 0. Proof. By the ultrametric inequality, the partial sums form a Cauchy sequence iff the b p 0. We can now do analysis in the p-adics. Definition 1. Let B be an open ball in Z p. A function f : B Z p is p-adic analytic if it is defined by a power series f(z) = a (z b 0 ) 0 for some b 0 B, with the power-series convergent for all z B. There are two natural ways to differentiate such a power series f. For simplicity, let s assume it s centered at zero and has radius of convergence at least 1 (namely it converges on all of Z p ). Then we may formally differentiate f via f(z) = a z ; f (z) = a z 1. Note that the radius of convergence of f will be at least that of f, as differentiating has made the coefficients no larger in p-adic absolute value. On the other hand, of s i 0, s i 0, one may define f via a difference quotient: f (z) = lim i f(z + s i ) f(z) s i. It s not hard to chec that these two definitions agree, say by looing at monomials and using continuity of evaluation. In particular, this implies that a p-adic analytic function is infinitely differentiable, and is constant if and only if its derivative is identically zero. This lets us do things lie tae Taylor series, recenter our analytic function, and so on. In particular, an analytic function with radius of convergence R about b 0 will be analytic with radius of convergence R about any b 1 with b 0 b 1 p R. We ll now prove a beautiful theorem of Strassmann, which will be the main technical tool in our proof of the Mahler-Solem-Lech theorem. This result is analogous to the fact that a holomorphic function on a connected region is identically zero if it is zero on a set with limit points. 3

4 Theorem 2 (Strassmann). Let f : B Z p be a p-adic analytic function. Then either f is identically zero, or has only finitely many zeros in B. Lemma 3. Z p is compact. Proof. Let {b } be a sequence of elements of Z p ; we wish to show there is a convergent subsequence. Let a i be the i-th digit of b. Then the a 1 tae some fixed value for infinitely many indices j ; among these j, the a 2j tae some value infinitely often, and so on. Collecting all of these values gives the decimal expansion of a limit point. Proof of Strassmann s theorem. Suppose f has infinitely many zeros, say f(b ) = 0. Then by our lemma above, the b have a limit point b. Let expand f about b via f(z) = a (z b). If f is not identically zero, some a 0; let a N be the first such coefficient. Then and so for z b p very small, we have that f(z) = (z b) N (a N + (z b)g(z)), (z b)g(z) p < a N p. In particular, f is non-zero in some small punctured dis about b. But this contradicts the fact that b was a limit point of the b. 4. Interlude: There are infinitely many primes Consider the following topology on the integers: A basis is given by sets of the form S(a, b) := az + b, a 0. Note that the S(a, b) are both open and closed, and any non-empty open set is infinite. Using the fact that this is a topology, we may prove the following beautiful theorem: Theorem 3. There are infinitely many primes. Proof. Assume the contrary. Then there are finitely may primes p i, and S(p i, 0) = Z \ {±1}. In particular, {±1} is open, which is a contradiction. i (This is of course just Euclid s proof in disguise.) Why do I mention this? Well, the topology I ve described on Z is exactly the subspace topology induced by the inclusion Z p Z p. This should suggest that p-adic analysis might have something to say about arithmetic progressions. Namely, let s imagine for a second that given an integer linear recurrence sequence a we could find a collection of p-adic analytic functions f i, 0 i m 1 such that f i (n) = nm + i for large n. Then each f i would either be identically zero, or would have only finitely many zeroes, which would complete the proof. The goal is to find such functions; I thin this is a natural goal, given that Z is dense in Z p, and we lie to extend functions (in this case a ) on dense subsets to their closures. 4

5 Let me remind you what we re trying to prove. 5. The Mahler-Solem-Lech Theorem Theorem 4 (Solem-Mahler-Lech). Let a n be a sequence defined by an integer linear recurrence. Then the set A := {n a n = 0} is the union of a finite set and a finite collection of (right)-infinite arithemtic progressions, i.e. sets of the form {qn + r n N}. Proof. Let s rewrite our problem to try to find f i as above. An integer linear recurrence sequence a may be written as a = A v, w for A an n n integer matrix, v, w vectors with integer entries (here v is the initial conditions; A is the transition matrix; and w pics out the top entry of a vector). Now choose some p not dividing the determinant of A; in particular A is invertible mod p. Now the group of invertible matrices mod p is finite, so in particular A m = 1 mod p for some m; this will eventually be the period of our arithmetic progressions. Let us write A m = I + pb for some matrix B with integer entries. Now let us construct our f i. For 0 i m 1, we have f i (n) = a mn+i = A mn A i v, w = (I + pb) n A i v, w. This is looing good! Expanding (1 + pb) n via the binomial formula, we find that f i (n) = p P (n) for some polynomials P in n with integer coefficients (these come from binomial coefficients). This power series maes sense as a p-adic analytic function convergent on all of Z p! But now we re done each f i has only finitely many zeros or is identically zero by Strassmann s theorem. This proof is due to Hansel. Note that we ve effectively bounded the period of the zero sets, namely by the order of GL n (F p ), where n is the length of the recurrence relation and p is the smallest prime not dividing det(a) above. It is not hard to generalize this argument to a similar result over any domain whose fraction field is characteristic zero. The theorem fails dramatically in finite characteristic. For example, consider over the field F p (t) the recurrence x n = (t + 1) n t n 1 defined by x n = (2x + 2)x n 1 (x 2 + 3x + 1)x n 2 + (x 2 + x)x n 3. Then the zero set is exactly the p-th powers. On the other hand, both the number and size of the exceptional zeros are mysterious from this proof. Evertse, Schlicewei, and Schmidt give a difficult bound on the number of exceptional zeros, (essentially by analyzing the f i ), but there is (as of 2007, and I assume now) no effective bound on their size. In particular, the following is open Question 1. Is there a computer program that decides whether or not an integer linear recurrence has any zeros? 5